All Exams Test series for 1 year @ ₹349 only
Question

When the potential difference across a resistive conductor of resistance R obeying ohm's law, is doubled, the rate of generation of heat will become

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
Four times

Heat Generation Rate Formula

The rate of heat generation in a resistive conductor is equivalent to the electrical power (P) dissipated by it. According to Ohm's law, the relationship between power, potential difference (\(V\)), and resistance (\(R\)) is given by:

\(P = \frac{V^2}{R}\)

This formula shows that the rate of heat generation (\(P\)) is directly proportional to the square of the potential difference (\(V^2\)) and inversely proportional to the resistance (\(R\)).

Power Change Calculation

Let the initial potential difference be \(V_1\) and the initial rate of heat generation be \(P_1\). Using the formula:

\(P_1 = \frac{V_1^2}{R}\)

The question states that the potential difference is doubled. Let the new potential difference be \(V_2\). Therefore:

\(V_2 = 2V_1\)

Let the new rate of heat generation be \(P_2\). Substituting \(V_2\) into the power formula:

\(P_2 = \frac{V_2^2}{R} = \frac{(2V_1)^2}{R}\)

Simplifying the expression for \(P_2\):

\(P_2 = \frac{4V_1^2}{R}\)

Now, compare \(P_2\) with the initial power \(P_1\):

\(P_2 = 4 \times \left( \frac{V_1^2}{R} \right) = 4P_1\)

Thus, when the potential difference is doubled, the rate of generation of heat becomes four times the original rate.

Was this answer helpful?

Similar Questions

  1. A bulb is connected across a battery of 10 V for 20 seconds. If a current of 2 A flows through it, calculate the heat energy generated by the bulb.
  2. An 8 Ω resistor generates 200 J of heat every second. What is the potential difference across the resistor?
  3. An electric bulb is connected to a 200 V generator. The current is 0.5 A. The power of the bulb is
  4. If Power = 100 watt and Resistance =4 ohms, then Current (I) =________?
  5. 450V bulb passes a current of 0.4A. Calculate the power in the lamp.

Important Questions from Power in Electric Circuits

  1. Two electric bulbs marked 25 W - 220 V and 100 W - 220 V are connected in series with 440 V supply. Which of the bulb will fuse?

  2. Two heaters are marked 200V,300W & 200V,600W. If the heaters are connected in series and the combination connected in series and the combination connected to a 200 V dc supply, then which option is true out of the following?

  3. Two electric bulbs are rated '$220$ V, $100$ W' and '$220$ V, $50$ W' respectively.
    If these two bulbs are connected in series across a $220$ V supply, what will be the power consumed by the $100$ W bulb?
  4. Three identical bulbs are connected in parallel to a battery of 6 V. If the current in the circuit is 0.6 A, the power dissipated by the battery is:
  5. A bulb is connected across a battery of 10 V for 20 seconds. If a current of 2 A flows through it, calculate the heat energy generated by the bulb.
Need Expert Advice?
Upcoming Exams
RRB Technician
October 06, 2026
RRB JE
October 27, 2026
RRB ALP
November 03, 2026
Test Series
RRB ALP img
Railways
RRB ALP 2026 Mock Test series
1035 Tests 1 Tests Free
1083 Attempts
4.3(239)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App